I am reading an Euclidean geometry text which, at some point, involved the behavior of a quantity of the form $\int_0^\infty x^n\, exp(-x)$. It says then that this quantity behaves like $o(1/x^2)$ when $x \rightarrow +\infty$.
I don't see why it is so. I mean, to me, we can expand $$x^n\, exp(-x)=x^n \, (1-x +\frac{x^2}{2!}-\frac{x^3}{3!}+...)$$ and while I can see that this could indeed converge (both intuitively from the $e^{-x}$ and from the fact it's an alternate series), I don't see how this reduced to an $o(1/x^2)$ behavior. If indeed we get an $o(1/x^2)$ then I agree that from Riemann's criteria, it would converge, but to get it in the first place is blurry to me. Any suggestion perhaps?
Hint:
Using the expansion $\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \ldots$ with $x > 0$,
$$e^x > \frac{x^{n+3}}{(n+3)!} \implies \frac{x^n e^{-x}}{1/x^2 } < \frac{(n+3)!}{x}$$